32g(g+1)=0

Solution for 32g(g+1)=0 equation:

32g(g+1)=0

We multiply parentheses

32g^2+32g=0

a = 32; b = 32; c = 0;
Δ = b2-4ac
Δ = 322-4·32·0
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-32}{2*32}=\frac{-64}{64}
=-1
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+32}{2*32}=\frac{0}{64}
=0
$